Problem Set #3
Math 471 – Real Analysis
Assignment: Chapter 3, #2, 3, 9, 10, 13
Clayton J. Lungstrum
September 19, 2012
Exercise 3.2
When b = 3 in Exercise 1, the expansion is called the triadic or ternary expansion of
x. Show that the Cantor set C consists of all x such that x has some triadic expansion for
which every ck is either 0 or 2.
be the Cantor-Lebesgue function: see p. 35.
that if x 2 C and x =
P1Let f (x)
P1 Show
1
k
k
c
3
,
where
each
c
is
either
0
or
2,
then
f
(x)
=
(
c
)2
.
k
k=1 k
k=1 2 k
Solution.
Let x 2 C, the Cantor set, with ternary expansion x = 0.c1 c2 c3 . . .. Then x 2 Cn , where
Cn is the nth stage of construction of the Cantor set, for all n
1. Therefore, at the first
stage of the construction, if x is in the leftmost interval, set c1 in the ternary expansion to 0;
if x is in the rightmost interval, set c1 = 2. At the next step, the interval x was in is divided
into 2 subintervals, x is in either the leftmost subinterval of the previous interval, or it is
in the rightmost subinterval of the previous interval. In the former case, set c2 = 0, in the
latter, c2 = 2. Continuing in this manner, we see that the Cantor set consists of all of those
numbers whose ternary expansion consists solely of 0’s and 2’s.
Conversely, suppose x has a ternary expansion
1
X
ck
3k
k=1
where each ck 2 {0, 2}. We will show x 2 C by induction. Clearly x 2 C0 since 0  x  1.
Next, if c1 = 0, then
1
x  (0.0222 . . .)3 = (0.1)3 = .
3
On the other hand, if c1 = 2, then
x
2
(0.2000 . . .)3 = (0.2)3 = ,
3
and hence, x 2 C1 . Now we approach the inductive step.
Assume x 2 Ck for some k 2 N. Then x is in any one of the 2k subintervals of Ck , each
of length 31k . Without loss of generality, say x 2 [a, b]. Since we remove the interior of the
middle-thirds, we have two disjoint closed intervals,


1
1
a, a + k+1
and
b
,b .
k+1
3
3
1
If ck+1 = 0, then x  a + 3k+1
and is therefore in the left interval. If ck+1 = 2, then
1
x
b 3k+1 , and then x would be in the right interval. Either way, x 2 Ck+1 . Thus, by
induction, we have shown x 2 C.
Combining both of these properties, we have that x 2 C if and only if its ternary expansion
consists of 0’s and 2’s only.
For the second part of the problem, let f (x) be the Cantor-Lebesgue function. Then
(n)
f (x) is continuous and thus f (x) = limn!1 f (ax ). By the definition of f (x), we have
1
(n)
(n)
f (x) = limk!1 fk (x) and fk (ax ) = fk+1 (ax ) for all k
in the textbook. Thus,
f (a(n)
x )
=
fm (a(n)
x )
=
n
X
⇥
fm (a(k)
x )
k=1
n, where fk (x) is defined here as
fm (a(k
x
1)
⇤
) ,
(0)
where fm (ax ) = 0. Then we have that this equals
n
X
ck
k=1
(n)
Thus, f (x) = limn!1 f (ax ) =
P1
ck
k
k=1 2 2 ,
2
2 k.
as desired.
Q.E.D.
2
Exercise 3.3
Construct a two-dimensional Cantor set in the unit square {(x, y) : 0  x, y  1} as
follows. Subdivide the square into nine equal parts and keep only the four closed corner
squares, removing the remaining region (which forms a cross). Then repeat this process in a
suitably scaled version for the remaining squares, ad infinitum. Show that the resulting set
is perfect, has plane measure zero, and equals C ⇥ C.
Solution.
Let us denote this two-dimensional Cantor set by C 2 , and let C02 be the unit square (the
first stage T
in the construction) while Ck2 is the k th stage in the construction. Thus, it is clear
2
that C 2 = 1
k=1 Ck . Note that we’re removing the interior of intervals in each iteration, thus
C 2 is the complement of an open set, thus is closed. Now we will show that each point is
a limit point. To that end, let A be a corner point of the closed square in Ck2 . Then notice
that A 2 C 2 since A 2 Cn2 for all n k. Using this fact, let P be a point in C 2 . Then, for
every k 0, there is a unique closed square containing P . Let Ak be a corner point di↵erent
from P in the k th stage of construction in the same square containing P . Then
p Pk is a limit
1
point of the sequence of corner points given by {Ak }k=1 since |P Ak | < 23 and each
corner point is in C 2 as established above. Thus, C 2 is closed and every point is a limit point,
therefore C 2 is a perfect set.
It’s clear that C 2 is measurable since it is the countable intersection of measurable sets
(established in class). Notice that there are 4k squares in each k th iteration, each of area 9 k .
k
Therefore, |Ck2 | = 49
, which tends to zero as k tends to infinity. From the monotonicity
of measurability, we see that |C 2 | = 0.
Our final step is to establish C 2 = C ⇥ C. To prove this, we’ll use mathematical induction.
Clearly, C02 = C0 ⇥ C0 . Then, assume that Ck2 = Ck ⇥ Ck for some k 0. Now, let Q = (x, y) 2
2
Ck2 . If Q 2 Ck+1
, then from construction, this means that ⇡x (x) is in either the first third or
the last third of the closed square (relative to the x-axis) in which it was contained in the
2
set Ck2 . A similar argument shows this to be true for y. However, if Q 2
/ Ck+1
, then either
the x or the y was in the middle third of its respective axis to be removed, i.e., ⇡x (x) 2
/ C or
⇡y (y) 2
/ C. Either way, Q 2
/ Ck ⇥ Ck . Hence, Ck2 = Ck ⇥ Ck , and it follows that
!
!
1
1
1
1
\
\
\
\
C2 =
Ck2 =
(C ⇥ C) =
C ⇥
C = C ⇥ C.
k=0
k=1
k=1
k=1
Q.E.D.
3
Exercise 3.9
P1
If {Ek }1
k=1 is a sequence of set with
k=1 |Ek | < 1, show that lim sup Ek (and so also
lim inf Ek ) has measure zero.
Solution.
Recall that if {Ek }1
k=1 is a sequence of sets, then
lim sup Ek =
1
\
j=1
1
[
Ek
k=j
!
.
For simplicity, let E = lim sup Ek . Then clearly
E✓
1
\
j=n
1
[
Ek
k=j
!
✓
1
[
Ek .
k=n
Thus, by monotonicity and countable subadditivity, we have
|E|e 
1
[
Ek
k=n
e

1
X
k=n
|Ek |e .
P
Now, P
since 1
N,
k=1 |Ek |e < 1, for any " > 0, we can choose N 2 N such that if n
then 1
|E
|
<
".
Thus,
applying
this
to
the
above
inequality,
for
every
"
>
0,
we
have
k e
k=n
|E|e < ", thus |E|e = 0. As we showed in class, E is then measurable and |E| = 0. Since
lim inf Ek ✓ lim sup Ek , by monotonicity, | lim inf Ek | = 0.
Q.E.D.
4
Exercise 3.10
If E1 and E2 are measurable, show that |E1 [ E2 | + |E1 \ E2 | = |E1 | + |E2 |.
Solution.
First, assume that |E1 | < 1 and |E2 | < 1, otherwise the result is trivial. Notice that
since E1 and E2 are measurable, we have E1 [ E2 measurable, thus
|E1 [ E2 | = |(E1 [ E2 ) \ E1 | + |(E1 [ E2 ) \ E1C | = |E1 | + |E2 \ E1C |.
Similarly, notice that since E2 is measurable, we have
|E2 | = |E2 \ E1 | + |E2 \ E1C |.
Now, taking the left-hand side of the first identity and adding it to the right-hand side of
the second identity, and similarly for the opposite sides, we have
|E1 [ E2 | + |E2 \ E1 | + |E2 \ E1C | = |E2 | + |E1 | + |E2 \ E1C |.
From both sides we can subtract |E2 \ E1C | and so we obtain the desired result.
Q.E.D.
5
Exercise 3.13
Motivated by (3.7), define the inner measure of E by |E|i = sup |F |, where the supremum
is taken over all closed subsets F of E. Show that |E|i  |E|e and if |E|e < 1, then E is
measurable if and only if |E|i = |E|e . [Use (3.22).]
Solution.
Let E ✓ Rn , let G be an open set containing E, and F ✓ E closed. Since G is open, it is
measurable, thus |G| = |G \ F | + |G \ F C | = |F | + |G F |. Thus, |F | = |G| |G F |, and
since Lebesgue measure is nonnegative, we clearly have |F ||  |G|. Since |F | is a lower bound
for |G|, it is clearly less than or equal to the greatest lower bound, i.e., |F |  inf |G| = |E|e .
Now we have on the other hand |E|e as an upper bound for |F |, thus it must be greater than
or equal to the least upper bound, i.e., |E|i = sup |F |  |E|e , and so we have the desired
inequality |E|i  |E|e .
Now suppose |E|e < 1 and that E is measurable. Then, given " > 0, there is a closed
set F 2 M with |E F |e < ". Since F 2 M,
|E|e = |E \ F |e + |E
This implies |E \ F |e > |E|e
F |e < |E \ F |e + ".
". Now, since |E|i
|E|i
|F |
|F | and F is measurable, we have
|F E|e = |E \ F |e .
| {z }
=0
Thus, we have
|E|e
|E|i > |E|e
".
Since " > 0 was arbitrary, we see that |E|i = |E|e .
Conversely, suppose |E|e < 1 and |E|i = |E|e . Then theorem (3.8) implies that there
exist sets of type F and G with the property that F ✓ E ✓ G, where F is F and G is
G , and |F | = |G|. Then E di↵ers only by a set of measure zero, and hence is measurable
by Theorem (3.28).
Q.E.D.
6